Write Raoult's law for non - volatile solute and volatile solvent and explain it.

Explanation

The vapour pressure of a solvent in solution is less than that of the pure solvent because the lowering of vapour presure depends only on the concentration of the solute particles.
A relation between vapour pressure of the solution, mole fraction and vapour pressure of the solvent, i.e., `p_(1)=x_(1).p_(1)^(0)`
The reduction in the vapour pressure of solvent `(Delta p_(1))` is given as : `p_(1)=x_(1).p_(1)^(0)`
It should be decrease in vapour presure =
`p_(1)^(0)-p_(1)=p_(1)^(0)-p_(1)^(0)x_(1)`
`1-x_(1)=x_(2)`
`Delta p_(1)=x_(2).p_(1)^(0)`
In a solution containing several non - volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
`(Delta p_(1))/(p_(1)^(0))=(p_(1)^(0)-p_(1))/(p_(1)^(0))=x_(2)`
So, relative lowering of vapour presure `((p_(1)^(0)-p_(1))/(p_(1)^(0)))` is equal to the mole fraction `(X_(2))` of the solute.
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))`
where, `n_(1)=` moles of solvent and
`n_(2)=` moles of solute
For dilute solutions `n2 lt lt n1`, hence neglecting n2 in the denominator we have,
`(p_(1)^(0)p_(1))/(p_(1)^(0))=(n_(2))/(n_(1))`
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xx M_(1))/(M_(2)xx w_(1))`
where, `w_(1)=` weight of solvent
`w_(2)=` weight of solute
`M_(1)=` molar weight of solvent
`M_(2)=` molar weight of solute.